Block Diagrams, StateVariable Models, and Simulation Methods


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1 5 C H A P T E R Block Diagram, StateVariable Model, and Simulation Method CHAPTER OUTLINE CHAPTER OBJECTIVES Part I. Model Form Tranfer Function and Block Diagram Model StateVariable Model 26 Part II. MATLAB Method StateVariable Method with MATLAB The MATLAB ode Function 279 Part III. Simulink Method Simulink and Linear Model Simulink and Nonlinear Model Chapter Review 38 Reference 39 Problem 39 When you have finihed thi chapter, you hould be able to. Decribe a dynamic model a a block diagram. 2. Derive ytem tranfer function from a block diagram. 3. Convert a differential equation model into tatevariable form. 4. Expre a linear tatevariable model in the tandard vectormatrix form. 5. Apply the, data, tfdata, eig, and initial function to analyze linear model. 6. Ue the MATLAB ode function to olve differential equation. 7. Ue Simulink to create imulation of dynamic model. Dynamic model derived from baic phyical principle can appear in everal form:. A a ingle equation (which i called the reduced form), 2. A a et of coupled firtorder equation (which i called the Cauchy or tatevariable form), and 3. A a et of coupled higherorder equation. 25
2 5. Tranfer Function and Block Diagram Model 25 In Chapter 2, 3, and 4 we analyzed the repone of a ingle equation, uch a mẍ + cẋ + kx= f (t), and et of coupled firtorder equation, uch a ẋ = 5x +7y, ẏ = 3x 9y + f (t), by firt obtaining the tranfer function and then uing the tranfer function to obtain a ingle, but higherorder equation. We alo obtained the repone of model that conit of a et of coupled higherorder equation. Each form ha it own advantage and diadvantage. We can convert one form into another, with differing degree of difficulty. In addition, if the model i linear, we can convert any of thee form into the tranfer function form or a vectormatrix form, each of which ha it own advantage. GUIDE TO THE CHAPTER Thi chapter ha three part. Part I i required to undertand Part II and III, but Part II and III are independent of each other. In Part I, which include Section 5. and 5.2, we introduce block diagram, which are baed on the tranfer function concept, and the tatevariable model form. The block diagram i a way of repreenting the dynamic of a ytem in graphical form. Block diagram will be ued often in the ubequent chapter to decribe dynamic ytem, and they are alo the bai for Simulink programming covered in Part III. An advantage of the tatevariable form i that it enable u to expre a linear model of any order in a tandard and compact way that i ueful for analyi and oftware application. In Chapter 2 we introduced the tf, tep, and lim function, which can olve model in tranfer function form. MATLAB ha a number of ueful function that are baed on the tatevariable model form. Thee function are covered in Part II, which include Section 5.3 and 5.4. Section 5.3 deal with linear model. While the analyi method of the previou chapter are limited to linear model, the tatevariable form i alo ueful for olving nonlinear equation. It i not alway poible or convenient to obtain the cloedform olution of a differential equation, and thi i uually true for nonlinear equation. Section 5.4 introduce MATLAB function that are ueful for olving nonlinear differential equation. Part III include Section 5.5 and 5.6 and introduce Simulink, which provide a graphical uer interface for olving differential equation. It i epecially ueful for olving problem containing nonlinear feature uch a Coulomb friction, aturation, and dead zone, becaue thee feature are very difficult to program with traditional programming method. In addition, it graphical interface might be preferred by ome uer to the more traditional programming method offered by the MATLAB olver covered in Part II. In Section 5.5 we begin with olving linear equation o that we can check the reult with the analytical olution. Section 5.6 cover Simulink method for nonlinear equation. PART I. MODEL FORMS 5. TRANSFER FUNCTIONS AND BLOCK DIAGRAM MODELS We have een that the complete repone of a linear ordinary differential equation (ODE) i the um of the free and the forced repone. For zero initial condition, the free repone i zero, and the complete repone i the ame a the forced repone.
3 252 CHAPTER 5 Block Diagram, StateVariable Model, and Simulation Method Thu, we can focu our analyi on the effect of only the input by taking the initial condition to be zero temporarily. When we have finihed analyzing the effect of the input, we can add to the reult the free repone due to any nonzero initial condition. The tranfer function i ueful for analyzing the effect of the input. Recall from Chapter 2 that the tranfer function T () i the tranform of the forced repone X () divided by the tranform of the input F(). T () = X () F() The tranfer function can be ued a a multiplier to obtain the forced repone tranform from the input tranform; that i, X () = T ()F(). The tranfer function i a property of the ytem model only. The tranfer function i independent of the input function and the initial condition. The tranfer function i equivalent to the ODE. If we are given the tranfer function we can recontruct the correponding ODE. For example, the tranfer function X () F() = correpond to the equation ẍ + 7ẋ + x = 5 f (t). Obtaining a tranfer function from a ingle ODE i traightforward, a we have een. Sometime, however, model have more than one input or more than one output. It i important to realize that there i one tranfer function for each inputoutput pair. If a model ha more than one input, the tranfer function for a particular output variable i the ratio of the tranform of the forced repone of that variable divided by the input tranform, with all the remaining input ignored (et to zero temporarily). For example, if the variable x i the output for the equation 5ẍ + 3ẋ + 4x = 6 f (t) 2g(t) then there are two tranfer function, X ()/F() and X ()/G(). Thee are X () F() = X () G() = We can obtain tranfer function from ytem of equation by firt tranforming the equation and then algebraically eliminating all variable except for the pecified input and output. Thi technique i epecially ueful when we want to obtain the repone of one or more of the dependent variable in the ytem of equation. For example, the tranfer function X ()/V () and Y ()/V () of the following ytem of equation: ẋ = 3x + 2y ẏ = 9y 4x + 3v(t) are and X () V () = Y () 3( + 3) = V ()
4 5. Tranfer Function and Block Diagram Model BLOCK DIAGRAMS We can ue the tranfer function of a model to contruct a viual repreentation of the dynamic of the model. Such a repreentation i a block diagram. Block diagram can be ued to decribe how ytem component interact with each other. Unlike a chematic diagram, which how the phyical connection, the block diagram how the caue and effect relation between the component, and thu help u to undertand the ytem dynamic. Block diagram can alo be ued to obtain tranfer function for a given ytem, for cae where the decribing differential equation are not given. In addition, a we will ee in Section 5.5, block diagram can be ued to develop imulation diagram for ue with computer tool uch a Simulink BLOCK DIAGRAM SYMBOLS Block diagram are contructed from the four baic ymbol hown in Figure 5..:. The arrow, which i ued to repreent a variable and the direction of the caueandeffect relation; 2. The block, which i ued to repreent the inputoutput relation of a tranfer function; 3. The circle, generically called a ummer, which repreent addition a well a ubtraction, depending on the ign aociated with the variable arrow; and 4. The takeoff point, which i ued to obtain the value of a variable from it arrow, for ue in another part of the diagram. The takeoff point doe not modify the value of a variable; a variable ha the ame value along the entire length of an arrow until it i modified by a circle or a block. You may think of a takeoff point a the tip of a voltmeter probe ued to meaure a voltage at a point on a wire. If the voltmeter i welldeigned, it will not change the value of the voltage it i meauring SOME SIMPLE BLOCK DIAGRAMS The implet block diagram i hown in Figure 5..b. Inide the block i the ytem tranfer function T (). The arrow going into the block repreent the tranform of the input, F(); the arrow coming out of the block repreent the tranform of the output, X (). Thu, the block diagram i a graphical repreentation of the caueandeffect relation operating in a particular ytem. A pecific cae i hown in Figure 5..2a in which the contant tranfer function K repreent multiplication and the block i called a multiplier or a gain block. The correponding equation in the time domain i x(t) = Kf(t). Another imple cae i hown in Figure 5..2b in which the tranfer function / repreent integration. The correponding equation in the time domain i x(t) = f (t) dt. Thu, uch a block i called an integrator. Note that thi relation correpond to the differential equation ẋ = f (t). X() (a) F() X() T() X() 5 T()F() (b) X() Z() 2 Y() Z() 5 X() 2 Y() (c) X() (d) X() Figure 5.. The four baic ymbol ued in block diagram.
5 254 CHAPTER 5 Block Diagram, StateVariable Model, and Simulation Method Figure 5..2 Two type of block. (a) Multiplier. (b) Integrator. Figure 5..3 Diagram repreenting the equation ẋ + 7x = f (t). F() K (a) X() F() (b) X() F() 7 X() F() 2 7 X() (a) (b) 5..4 EQUIVALENT BLOCK DIAGRAMS Figure 5..3 how how more than one diagram can repreent the ame model, which in thi cae i ẋ + 7x = f (t). The tranfer function i X ()/F() = /( + 7), and the correponding diagram i hown in part (a) of the figure. However, we can rearrange the equation a follow: ẋ = f (t) 7x or x = [ f (t) 7x] dt which give X () = [F() 7X ()] In thi arrangement the equation correpond to the diagram hown in part (b) of the figure. Note the ue of the takeoff point to feed the variable X () to the multiplier. The circle ymbol i ued to repreent addition of the variable F() and ubtraction of 7X (). The diagram how how ẋ, the rate of change of x, i affected by x itelf. Thi i hown by the path from X () through the multiplier block to the ummer, which change the ign of 7X (). Thi path i called a negative feedback path or a negative feedback loop SERIES ELEMENTS AND FEEDBACK LOOPS Figure 5..4 how two common form found in block diagram. In part (a) the two block are aid to be in erie. It i equivalent to the diagram in part (b) becaue we may write B() = T ()F() X () = T 2 ()B() Thee can be combined algebraically by eliminating B() to obtain X () = T () T 2 ()F(). Note that block diagram obey the rule of algebra. Therefore, any rearrangement permitted by the rule of algebra i a valid diagram. Figure 5..4 (a) and (b) Simplification of erie block. (c) and (d) Simplification of a feedback loop. F() B() X() T () T 2 () (a) F() A() G() X() 2 B() H() F() F() T ()T 2 () (b) G() G()H() X() X() (c) (d)
6 5. Tranfer Function and Block Diagram Model 255 Figure 5..4c how a negative feedback loop. From the diagram, we can obtain the following. A() = F() B() B() = H()X () X () = G()A() We can eliminate A() and B() to obtain G() X () = F() (5..) + G()H() Thi i a ueful formula for implifying a feedback loop to a ingle block REARRANGING BLOCK DIAGRAMS Now conider the econdorder model ẍ + 7ẋ + x = f (t). The tranfer function i X ()/F() = /( ), and the implet diagram for thi model i hown in Figure 5..5a. However, to how how x and ẋ affect the dynamic of the ytem, we can contruct a diagram that contain the appropriate feedback path for x and ẋ. To do thi, rearrange the equation by olving for the highet derivative. The tranformed equation i X () = ẍ = f (t) 7ẋ x ( ) {F() 7[X()] X ()} With thi arrangement we can contruct the diagram hown in Figure 5..5b. Recall that X() repreent ẋ. The term within the pair of curly brace i the output of the ummer and the input to the leftmot integrator. The output of thi integrator i hown within the outermot pair of parenthee and i the input to the rightmot integrator. We may ue two ummer intead of one, and rearrange the diagram a hown in Figure 5..5c. Thi form how more clearly the negative feedback loop aociated with the derivative ẋ. Referring to Figure 5..3, we ee that we may replace thi inner loop with it equivalent tranfer function /( + 7). The reult i hown in Figure 5..5d, which diplay only the feedback loop aociated with x. Figure 5..5 Diagram repreenting the equation ẍ + 7ẋ + x = f (t). F() 2 7 X() F() X() 2 2 X() 7 (a) (b) F() 2 2 X() F() 2 7 X() 7 (c) (d)
7 256 CHAPTER 5 Block Diagram, StateVariable Model, and Simulation Method Two important point can be drawn from thee example.. More than one correct diagram can be drawn for a given equation; the deired form of the diagram depend on what information we want to diplay. 2. The form of the reulting diagram depend on how the equation i arranged. A ueful procedure for contructing block diagram i to firt olve for the highet derivative of the dependent variable; the term on the right ide of the reulting equation repreent the input to an integrator block. It i important to undertand that the block diagram i a picture of the algebraic relation obtained by applying the Laplace tranform to the differential equation, auming that the initial condition are zero. Therefore, a number of different diagram can be contructed for a given et of equation and they will all be valid a long a the algebraic relation are correctly repreented. Block diagram are epecially ueful when the model conit of more than one differential equation or ha more than one input or output. For example, conider the model ẋ = 3y + f (t) ẏ = 5y + 4x + g(t) which ha two input, f (t) and g(t). Suppoe we are intereted in the variable y a the output. Then the diagram in Figure 5..6a i appropriate. Notice that it how how y affect itelf through the feedback loop with the gain of 3, by firt affecting x. Uually we try to place the output variable on the right ide of the diagram, with it arrow pointing to the right. We try to place one input on the left ide with it arrow point to the right, with a econd input, if any, placed at the top of the diagram. The diagram hown in Figure 5..6a follow thee convention, which have been etablihed to make it eaier for other to interpret your diagram. Jut a in the Englih language we read from left to right, o the main flow of the caue and effect in a diagram (from input to output) hould be from left to right if poible. If intead, we chooe the output to be x, then Figure 5..6b i more appropriate. Figure 5..6 A diagram with two input. G() F() X() Y() 3 (a) 5 F() G() Y() 4 (b) X()
8 5. Tranfer Function and Block Diagram Model TRANSFER FUNCTIONS FROM BLOCK DIAGRAMS Sometime we are given a block diagram and aked to find either the ytem tranfer function or it differential equation. There are everal way to approach uch a problem; the appropriate method depend partly on the form of the diagram and partly on peronal preference. The following example illutrate the proce. Serie Block and Loop Reduction EXAMPLE 5.. Problem Determine the tranfer function X ()/F() for the ytem whoe diagram i hown in Figure 5..7a. Solution When two block are connected by an arrow, they can be combined into a ingle block that contain the product of their tranfer function. The reult i hown in part (b) of the figure. Thi property, which i called the erie or cacade property, i eaily demontrated. In term of the variable X (), Y (), and Z() hown in the diagram, we can write X () = Y () Y () = Z() Eliminating Y () we obtain X () = Z() Thi give the diagram in part (b) of the figure. So we ee that combining two block in erie i equivalent to eliminating the intermediate variable Y () algebraically. To find the tranfer function X ()/F(), we can write the following equation baed on the diagram in part (b) of the figure: X () = Z() Z() = F() 8X () ( + 6)( + 2) Eliminating Z() from thee equation give the tranfer function X () F() = F() 2 Z() 6 (a) 8 Y() 2 X() F() 2 Z() ( 6)( 2) (b) 8 X() Figure 5..7 An example of erie combination and loop reduction. Uing Integrator Output EXAMPLE 5..2 Problem Determine the model for the output x for the ytem whoe diagram i hown in Figure F() W() Y() G() X() Figure 5..8 Diagram for Example
9 258 CHAPTER 5 Block Diagram, StateVariable Model, and Simulation Method Solution The input to an integrator block / i the time derivative of the output. Thu, by examining the input to the two integrator hown in the diagram we can immediately write the timedomain equation a follow. ẋ = g(t) + y ẏ = 7w 3x w = f (t) 4x We can eliminate the variable w from the lat two equation to obtain ẏ = 7 f (t) 3x. Thu, the model in differential equation form i ẋ = g(t) + y ẏ = 7 f (t) 3x To obtain the model in tranfer function form we firt tranform the equation. X() = G() + Y () Y() = 7F() 3X () Then we eliminate Y () algebraically to obtain 7 X () = F() G() There are two tranfer function, one for each inputoutput pair. They are X () F() = X () G() = Sometime, we need to obtain the expreion not for jut the output variable, but alo for ome internal variable. The following example illutrate the required method. EXAMPLE 5..3 Deriving Expreion for Internal Variable Problem Derive the expreion for C(), E(), and M() in term of R() and D() for the diagram in Figure Solution Start from the righthand ide of the diagram and work back to the left until all block and comparator are accounted for. Thi give C() = 7 [M() D()] () + 3 M() = K E() (2) 4 + E() = R() C() (3) Figure 5..9 Block diagram for Example R() 2 E() K 4 D() M() C()
10 5. Tranfer Function and Block Diagram Model 259 Multiply both ide of equation () by + 3 to clear fraction, and ubtitute M() and E() from equation (2) and (3). ( + 3)C() = 7M() 7D() K = 7 E() 7D() = 7K [R() C()] 7D() Multiply both ide by 4 + to clear fraction, and olve for C() to obtain: 7K C() = K R() 7(4 + ) D() (4) K The characteritic polynomial i found from the denominator of either tranfer function. It i K. The equation for E() i E() = R() C() 7K = R() K R() + 7(4 + ) K D() = K R() + 7(4 + ) K D() Becaue can be factored a (4 + )( + 3), the equation for M() can be expreed a M() = K 4 + E() = K [ ] (4 + )( + 3) K R() + 7(4 + ) K D() K ( + 3) = K R() + 7K K D() Note the cancellation of the term 4 +. You hould alway look for uch cancellation. Otherwie, the denominator of the tranfer function can appear to be of higher order than the characteritic polynomial BLOCK DIAGRAM ALGEBRA USING MATLAB MATLAB can be ued to perform block diagram algebra if all the gain and tranfer function coefficient have numerical value. You can combine block in erie or in feedback loop uing the erie and feedback function to obtain the tranfer function and the tatepace model. If the LTI model y and y2 repreent block in erie, their combined tranfer function can be obtained by typing y3 = erie(y,y2). A imple gain need not be converted to a LTI model, and doe not require the erie function. For example, if the firt ytem i a imple gain K, ue the multiplication ymbol * and enter y3 = K*y2 If the LTI model y2 i in a negative feedback loop around the LTI model y, then enter y3 = feedback(y,y2) to obtain the LTI model of the cloedloop ytem.
11 26 CHAPTER 5 Block Diagram, StateVariable Model, and Simulation Method Figure 5.. A typical block diagram. F() X() If the feedback loop i poitive, ue the yntax y3 = feedback(y,y2,+) If you need to obtain the numerator and denominator of the cloedloop tranfer function, you can ue the tfdata function and enter [num,den] = tfdata(y3,'v') You can then find the characteritic root by entering root(den) For example, to find the tranfer function X ()/F() correponding to the block diagram hown in Figure 5.., you enter y=tf(,[,]); y2=feedback(y,7); y3=erie(y,y2); y4=feedback(y3,); [num,den]=tfdata(y4,'v') The reult i num = [,, ] and den = [, 7, ], which correpond to X () F() = STATEVARIABLE MODELS Model that conit of coupled firtorder differential equation are aid to be in tatevariable form. Thi form, which i alo called the Cauchy form, ha an advantage over the reduced form, which conit of a ingle, higherorder equation, becaue it allow a linear model to be expreed in a tandard and compact way that i ueful for analyi and for oftware application. Thi repreentation make ue of vector and matrix notation. In thi ection, we will how how to obtain a model in tatevariable form and how to expre tatevariable model in vectormatrix notation. In Section 5.3 we how how to ue thi notation with MATLAB. Conider the econdorder equation 5ÿ + 7ẏ + 4y = f (t) Solve it for the highet derivative: ÿ = 5 f (t) 4 5 y 7 5 ẏ Now define two new variable, x and x 2, a follow: x = y and x 2 =ẏ. Thi implie that ẋ = x 2 and ẋ 2 = 5 f (t) 4 5 x 7 5 x 2
12 5.2 StateVariable Model 26 Thee two equation, called the tate equation, are the tatevariable form of the model, and the variable x and x 2 are called the tate variable. The general mapringdamper model ha the following form: mẍ + cẋ + kx = f (5.2.) If we define new variable x and x 2 uch that x = x x 2 =ẋ thee imply that ẋ = x 2 (5.2.2) Then we can write the model (5.2.) a: mẋ 2 + cx 2 + kx = f. Next olve for ẋ 2 : ẋ 2 = m ( f kx cx 2 ) (5.2.3) Equation (5.2.2) and (5.2.3) contitute a tatevariable model correponding to the reduced model (5.2.). The variable x and x 2 are the tate variable. If (5.2.) repreent a mapringdamper ytem, the tatevariable x decribe the ytem potential energy kx 2/2, which i due to the pring, and the tatevariable x 2 decribe the ytem kinetic energy mx2 2 /2, which i due to the ma. Although here we have derived the tate variable model from the reduced form, tatevariable model can be derived from baic phyical principle. Chooing a tate variable thoe variable that decribe the type of energy in the ytem ometime help to derive the model (note that k and m are alo needed to decribe the energie, but thee are parameter, not variable). The choice of tate variable i not unique, but the choice mut reult in a et of firtorder differential equation. For example, we could have choen the tate variable to be x = x and x 2 = mẋ, which i the ytem momentum. In thi cae the tatevariable model would be ẋ = m x 2 ẋ 2 = f c m x 2 kx StateVariable Model of a TwoMa Sytem EXAMPLE 5.2. Problem Conider the twoma ytem dicued in Chapter 4 (and hown again in Figure 5.2.). Suppoe the parameter value are m = 5, m 2 = 3, c = 4, c 2 = 8, k =, and k 2 = 4. The equation of motion are 5ẍ + 2ẋ + 5x 8ẋ 2 4x 2 = () Put thee equation into tatevariable form. 3ẍ 2 + 8ẋ 2 + 4x 2 8ẋ 4x = f (t) (2) Figure 5.2. A twoma ytem. k c m x Solution Uing the ytem potential and kinetic energie a a guide, we ee that the diplacement x and x 2 decribe the ytem potential energy and that the velocitie ẋ and ẋ 2 decribe the ytem kinetic energy. That i k 2 c 2 m 2 x 2 PE = 2 k x k 2(x x 2 ) 2 f
13 262 CHAPTER 5 Block Diagram, StateVariable Model, and Simulation Method and KE = 2 m ẋ m 2ẋ 2 2 Thi indicate that we need four tate variable. (Another way to ee that we need four variable i to note that the model conit of two coupled econdorder equation, and thu i effectively a fourthorder model.) Thu, we can chooe the tate variable to be x x 2 x 3 =ẋ x 4 =ẋ 2 (3) Thu, two of the tate equation are ẋ = x 3 and ẋ 2 = x 4. The remaining two equation can be found by olving equation () and (2) for ẍ and ẍ 2, noting that ẍ =ẋ 3 and ẍ 2 =ẋ 4, and uing the ubtitution given by equation (3). ẋ 3 = 5 ( 2x 3 5x + 8x 4 + 4x 2 ) ẋ 4 = 3 [ 8x 4 4x 2 + 8x 3 + 4x + f (t)] Note that the lefthand ide of the tate equation mut contain only the firtorder derivative of each tate variable. Thi i why we divided by 5 and 3, repectively. Note alo that the righthand ide mut not contain any derivative of the tate variable. Failure to oberve thi retriction i a common mitake. Now lit the four tate equation in acending order according to their lefthand ide, after rearranging the righthand ide o that the tate variable appear in acending order from left to right. Thee are the tate equation in tandard form. ẋ = x 3 (4) ẋ 2 = x 4 (5) ẋ 3 = 5 ( 5x + 4x 2 2x 3 + 8x 4 ) (6) ẋ 4 = 3 [4x 4x 2 + 8x 3 8x 4 + f (t)] (7) 5.2. VECTORMATRIX FORM OF STATEVARIABLE MODELS Vectormatrix notation enable u to repreent multiple equation a a ingle matrix equation. For example, conider the following et of linear algebraic equation. 2x + 9x 2 = 5 (5.2.4) 3x 4x 2 = 7 (5.2.5) The term matrix refer to an array with more than one column and more than one row. A column vector i an array having only one column. A row vector ha only one row. A matrix i an arrangement of number and i not the ame a a determinant, which can be reduced to a ingle number. Multiplication of a matrix having two row and two column (a (2 2) matrix) by a column vector having two row and one column (a (2 ) vector) i defined a follow: [ ][ ] [ ] a a 2 x a x = + a 2 x 2 (5.2.6) a 2 a 22 x 2 a 2 x + a 22 x 2
14 5.2 StateVariable Model 263 Thi definition i eaily extended to matrice having more than two row or two column. In general, the reult of multiplying an (n n) matrix by an (n ) vector i an (n ) vector. Thi definition of vectormatrix multiplication require that the matrix have a many column a the vector ha row. The order of the multiplication cannot be revered (vectormatrix multiplication doe not have the commutative property). Two vector are equal if all their repective element are equal. Thu we can repreent the et (5.2.4) and (5.2.5) a follow: [ ][ ] [ ] 2 9 x 5 = (5.2.7) x 2 We uually repreent matrice and vector in boldface type, with matrice uually in upper cae letter and vector in lowercae, but thi i not required. Thu we can repreent the et (5.2.7) in the following compact form. Ax = b (5.2.8) where we have defined the following matrice and vector: [ ] [ ] [ ] 2 9 x 5 A = x = b = The matrix A correpond in an ordered fahion to the coefficient of x and x 2 in (5.2.4) and (5.2.5). Note that the firt row in A conit of the coefficient of x and x 2 on the lefthand ide of (5.2.4), and the econd row contain the coefficient on the lefthand ide of (5.2.5). The vector x contain the variable x and x 2, and the vector b contain the righthand ide of (5.2.4) and (5.2.5). x 2 VectorMatrix Form of a SingleMa Model EXAMPLE Problem Expre the mapringdamper model (5.2.2) and (5.2.3) a a ingle vectormatrix equation. Thee equation are ẋ = x 2 ẋ 2 = m f (t) k m x c m x 2 Solution The equation can be written a one equation a follow: [ ] ẋ = [ ] ẋ 2 k m c x + f (t) x 2 m m In compact form thi i ẋ = Ax + B f (t) where A = k m c B = m m x = [ x x 2 ]
15 264 CHAPTER 5 Block Diagram, StateVariable Model, and Simulation Method EXAMPLE VectorMatrix Form of the TwoMa Model Problem Expre the tatevariable model of Example 5.2. in vectormatrix form. The model i ẋ = x 3 ẋ 2 = x 4 Solution In vectormatrix form thee equation are ẋ 3 = 5 ( 5x + 4x 2 2x 3 + 8x 4 ) ẋ 4 = 3 [4x 4x 2 + 8x 3 8x 4 + f (t)] ẋ = Ax + B f (t) where and A = x x x = x 2 x 3 = x 2 ẋ x 4 ẋ 2 B = STANDARD FORM OF THE STATE EQUATION We may ue any ymbol we chooe for the tate variable and the input function, although the common choice i x i for the tate variable and u i for the input function. The tandard vectormatrix form of the tate equation, where the number of tate variable i n and the number of input i m, i ẋ = Ax + Bu (5.2.9) where the vector x and u are column vector containing the tate variable and the input, if any. The dimenion are a follow: The tate vector x i a column vector having n row. The ytem matrix A i a quare matrix having n row and n column. The input vector u i a column vector having m row. The control or input matrix B ha n row and m column. In our example thu far there ha been only one input, and for uch cae the input vector u reduce to a calar u. The tandard form, however, allow for more than one input function. Such would be the cae in the twoma model if external force f and f 2 are applied to the mae.
16 5.2 StateVariable Model THE OUTPUT EQUATION Some oftware package and ome deign method require you to define an output vector, uually denoted by y. The output vector contain the variable that are of interet for the particular problem at hand. Thee variable are not necearily the tate variable, but might be ome combination of the tate variable and the input. For example, in the mapring model, we might be intereted in the total force f kx cẋ acting on the ma, and in the momentum mẋ. In thi cae, the output vector ha two element. If the tate variable are x = x and x 2 =ẋ, the output vector i [ ] [ ] [ ] y f kx cẋ f kx cx y = = = 2 y 2 mẋ mx 2 or y = [ y y 2 ] = [ k c m ][ x x 2 ] [ ] + f = Cx + D f where and C = [ ] k c m [ ] D = Thi i an example of the general form: y = Cx + Du. The tandard vectormatrix form of the output equation, where the number of output i p, the number of tate variable i n, and the number of input i m, i y = Cx + Du (5.2.) where the vector y contain the output variable. The dimenion are a follow: The output vector y i a column vector having p row. The tate output matrix C ha p row and n column. The control output matrix D ha p row and m column. The matrice C and D can alway be found whenever the choen output vector y i a linear combination of the tate variable and the input. However, if the output i a nonlinear function, then the tandard form (5.2.) doe not apply. Thi would be the cae, for example, if the output i choen to be the ytem kinetic energy: KE = mx 2 2 /2. The Output Equation for a TwoMa Model EXAMPLE Problem Conider the twoma model of Example a) Suppoe the output are x and x 2. Determine the output matrice C and D. b) Suppoe the output are (x 2 x ), ẋ 2, and f. Determine the output matrice C and D.
17 266 CHAPTER 5 Block Diagram, StateVariable Model, and Simulation Method Solution a. In term of the z vector, z = x and z 3 = x 2. We can expre the output vector y a follow. Thu y = [ x ] = x 2 C = [ [ ] ] x x 2 x 3 x 4 D = [ ] + f [ ] b. Here the output are y = x 2 x, y 2 =ẋ 2 = x 4, and y 3 = f. Thu we can expre the output vector a follow: y = x 2 x x 4 = x x 2 x f 3 + f x 4 Thu C = D = TRANSFERFUNCTION VERSUS STATEVARIABLE MODELS The deciion whether to ue a reducedform model (which i equivalent to a tranferfunction model) or a tatevariable model depend on many factor, including peronal preference. In fact, for many application both model are equally effective and equally eay to ue. Application of baic phyical principle ometime directly reult in a tatevariable model. An example i the following twoinertia fluidclutch model derived in Chapter 4. I d ω d = T d c(ω d ω ) I ω = T + c(ω d ω ) The tate and input vector are [ ] [ ] ωd Td x = u = ω T The ytem and input matrice are c c I d I d A = c c I I B = I d I For example, thi form of the model i eaier to ue if you need to obtain only numerical value or a plot of the tep repone, becaue you can directly ue the MATLAB function tep(a,b,c,d), to be introduced in Section 5.3. However, if you need to obtain the tep repone a a function, it might be eaier to convert the model to tranfer function
18 5.2 StateVariable Model 267 form and then ue the Laplace tranform to obtain the deired function. To obtain the tranfer function from the tatevariable model, you may ue the MATLAB function tf(y), a hown in Section 5.3. The MATLAB function cited require that all the model parameter have pecified numerical value. If, however, you need to examine the effect of a ytem parameter, ay the damping coefficient c in the clutch model, then it i perhap preferable to convert the model to tranfer function form. In thi form, you can examine the effect of c on ytem repone by examining numerator dynamic and the characteritic equation. You can alo ue the initial and finalvalue theorem to invetigate the repone MODEL FORMS HAVING NUMERATOR DYNAMICS Note that if you only need to obtain the free repone, then the preence of input derivative or numerator dynamic in the model i irrelevant. For example, the free repone of the model 5 d3 y dt + y 3 3d2 dt + 7dy + 6y = 4df 2 dt dt + 9 f (t) i identical to the free repone of the model 5 d3 y dt + y 3 3d2 dt + 7dy 2 dt + 6y = which doe not have any input. A tatevariable model for thi equation i eaily found to be x = y x 2 =ẏ x 3 = ÿ ẋ = x 2 ẋ 2 = x 3 ẋ 3 = 6 5 x 7 5 x x 3 The free repone of thi model can be eaily found with the MATLAB initial function to be introduced in the next ection. For ome application you need to obtain a tatevariable model in the tandard form. However, in the tandard tatevariable form ẋ = Ax + Bu there i no derivative of the input u. When the model ha numerator dynamic or input derivative, the tate variable are not o eay to identify. When there are no numerator dynamic you can alway obtain a tatevariable model in tandard form from a tranferfunction or reducedform model whoe dependent variable i x by defining x = x, x 2 = ẋ, x 3 = ẍ, and o forth. Thi wa the procedure followed previouly. Note that the initial condition x (), x 2 (), and x 3 () are eaily obtained from the given condition x(), ẋ(), and ẍ(); that i, x () = x(), x 2 () =ẋ(), and x 3 () = ẍ(). However, when numerator dynamic are preent, a different technique mut be ued, and the initial condition are not a eaily related to the tate variable. We now give two example of how to obtain a tatevariable model when numerator dynamic exit. Numerator Dynamic in a FirtOrder Sytem EXAMPLE Problem Conider the tranfer function model Z() U() = ()
19 268 CHAPTER 5 Block Diagram, StateVariable Model, and Simulation Method Thi correpond to the equation ż + 2z = 5 u + 3u (2) Note that thi equation i not in the tandard form ż = Az+ Bu becaue of the input derivative u. Demontrate two way of converting thi model to a tatevariable model in tandard form. Solution a. One way of obtaining the tatevariable model i to divide the numerator and denominator of equation () by. Z() U() = 5 + 3/ (3) + 2/ The objective i to obtain ainthedenominator, which i then ued to iolate Z() a follow: Z() = 2 Z() + 5U() + 3 U() = [3U() 2Z()] + 5U() The term within quare bracket multiplying / i the input to an integrator, and the integrator output can be elected a a tatevariable x. Thu, where Thi give with the output equation Z() = X () + 5U() X () = [3U() 2Z()] = {3U() 2 [X () + 5U()]} = [ 2X () 7U()] ẋ = 2x 7u (4) z = x + 5u (5) Thi fit the tandard form (5.2.9) and (5.2.), with A = 2, B = 7, y = z, C =, and D = 5. Preumably we are given the initial condition z(), but to olve equation (4) we need x(). Thi can be obtained by olving equation (5) for x, x = z 5u, and evaluating it at t = : x() = z() 5u(). We ee that x() = z() if u() =. b. Another way i to write equation () a Z() = (5 + 3) U() (6) + 2 and define the tatevariable x a follow: X () = U() (7) + 2 Thu, and the tate equation i X() = 2X () + U() (8) ẋ = 2x + u (9)
20 5.2 StateVariable Model 269 To find the output equation, note that Z() = (5 + 3) U() = (5 + 3)X () = 5X() + 3X () + 2 Uing equation (8) we have and thu the output equation i Z() = 5[ 2X () + U()] + 3X () = 7X () + 5U() z = 7x + 5u () The initial condition x() i found from equation () to be x() = [5u() z()]/7 = z()/7ifu() =. Although the model coniting of equation (9) and () look different than that given by equation (4) and (5), they are both in the tandard form and are equivalent, becaue they were derived from the ame tranfer function. Thi example point out that there i no unique way to derive a tatevariable model from a tranfer function. It i important to keep thi in mind becaue the tatevariable model obtained from the MATLAB data(y) function, to be introduced in the next ection, might not be the one you expect. The tatevariable model given by MATLAB i ẋ = 2x + 2u, z = 3.5x + 5u. Thee value correpond to equation () being written a Z() = U() = ( ) + 2 and defining x a the term within the quare bracket; that i, X () = 2U() + 2 [ 2U() + 2 ] The order of the ytem, and therefore the number of tate variable required, can be found by examining the denominator of the tranfer function. If the denominator polynomial i of order n, then n tate variable are required. Frequently a convenient choice i to elect the tate variable a the output of integration (/), a wa done in Example Numerator Dynamic in a SecondOrder Sytem EXAMPLE Problem Obtain a tatevariable model for X () U() = () Relate the initial condition for the tate variable to the given initial condition x() and ẋ(). Solution Divide by 5 2 to obtain ainthedenominator. 7 X () U() =
21 27 CHAPTER 5 Block Diagram, StateVariable Model, and Simulation Method Ue the in the denominator to olve for X (). ( 7 X () = ) ( 4 5 U() ) 5 2 X () = { 4 5 X () U() + [ 7 5 U() 7 ]} 5 X () (2) Thi equation how that X () i the output of an integration. Thu x can be choen a a tatevariable x. Thu, X () = X () The term within quare bracket in (2) i the input to an integration, and thu the econd tate variable can be choen a X 2 () = [ 7 5 U() 7 ] 5 X () = [ 7 5 U() 7 ] 5 X () (3) Then from equation (2) X () = [ 4 5 X () + 4 ] 5 U() + X 2() (4) The tate equation are found from (3) and (4). ẋ = 4 5 x + x u (5) ẋ 2 = 7 5 x u (6) and the output equation i x = x. The matrice of the tandard form are [ ] ] 4 5 A = B = 7 5 [ C = [ ] D = [] Note that the tate variable obtained by thi technique do not alway have traightforward phyical interpretation. If the model mẍ + cẋ + kx = c u + ku repreent a mapringdamper ytem with a diplacement input u with m = 5, c = 4, k = 7, the variable x 2 i the integral of the pring force k(u x), divided by the ma m. Thu, x 2 i the acceleration of the ma due to thi force. Sometime convenient phyical interpretation of the tate variable are acrificed to obtain pecial form of the tate equation that are ueful for analytical purpoe. Uing equation (5) and (6), we need to relate the value of x () and x 2 () to x() and ẋ(). Becaue x wa defined to be x = x, we ee that x () = x(). To find x 2 (), we olve the firt tate equation, equation (5), for x 2. Thi give Thu if u() =, x 2 =ẋ (x u) x 2 () =ẋ () [x () u()] =ẋ() + 4 [x() u()] 5 x 2 () =ẋ() x()
22 5.3 StateVariable Method with MATLAB 27 Table 5.2. A tatevariable form for numerator dynamic. Tranfer function model: Statevariable model: where Uual cae: where Y () U() = β n n + β n n + +β + β n + α n n + +α + α ẋ = γ n u α n x + x 2 ẋ 2 = γ n 2 u α n 2 x + x 3. ẋ j = γ n j u α n j x + x j+, j =, 2,...,n. ẋ n = γ u α x y = β n u + x γ i = β i α i β n If u() = u() = =, then x i () = y (i ) () + α n y (i 2) () + +α n i+ y() i =, 2,...,n y (i) () = di y dt i t= The method of the previou example can be extended to the general cae where the tranfer function i Y () U() = β n n + β n n + +β + β (5.2.) n + α n n + +α + α The reult are hown in Table The detail of the derivation are given in [Palm, 986]. PART II. MATLAB METHODS 5.3 STATEVARIABLE METHODS WITH MATLAB The MATLAB tep, impule, and lim function, treated in Section 2.9, can alo be ued with tatevariable model. However, the initial function, which compute the free repone, can be ued only with a tatevariable model. MATLAB alo provide function for converting model between the tatevariable and tranfer function form. Recall that to create an LTI object from the reduced form 5ẍ + 7ẋ + 4x = f (t) (5.3.) or the tranfer function form X () F() = (5.3.2) you ue the tf(num,den) function by typing: y = tf(, [5, 7, 4]); The reult, y, i the LTI object that decribe the ytem in the tranfer function form. The LTI object y2 in tranfer function form for the equation 8 d3 x dt x 3 3d2 dt + 5dx 2 dt + 6x = f 4d2 dt 2 + 3df dt + 5 f (5.3.3)
23 272 CHAPTER 5 Block Diagram, StateVariable Model, and Simulation Method i created by typing y2 = tf([4, 3, 5],[8, 3, 5, 6]); 5.3. LTI OBJECTS AND THE (A,B,C,D) FUNCTION To create an LTI object from a tate model, you ue the (A,B,C,D) function, where tand for tate pace. The matrix argument of the function are the matrice in the following tandard form of a tate model: ẋ = Ax + Bu (5.3.4) y = Cx + Du (5.3.5) where x i the vector of tate variable, u i the vector of input function, and y i the vector of output variable. For example, to create an LTI object in tatemodel form for the ytem decribed by ẋ = x 2 ẋ 2 = 5 f (t) 4 5 x 7 5 x 2 where x i the deired output, the required matrice are [ ] [ ] A = 4 7 B = In MATLAB you type A = [, ; 4/5, 7/5]; B = [; /5]; C = [, ]; D = ; y3 = (A,B,C,D); C = [ ] D = THE (y) AND data(y) FUNCTIONS An LTI object defined uing the tf function can be ued to obtain an equivalent tate model decription of the ytem. To create a tate model for the ytem decribed by the LTI object y created previouly in tranfer function form, you type (y). You will then ee the reulting A, B, C, and D matrice on the creen. To extract and ave the matrice a A, B, C, and D (to avoid overwriting the matrice from the econd example here), ue the data function a follow. [A, B, C, D] = data(y); The reult are [ ].4.8 A = [ ].5 B = C = [.4] D = [ ]
24 5.3 StateVariable Method with MATLAB 273 which correpond to the tate equation: ẋ =.4x.8x f (t) ẋ 2 = x and the output equation y =.4x RELATING STATE VARIABLES TO THE ORIGINAL VARIABLES When uing data to convert a tranfer function form into a tate model, note that the output y will be a calar that i identical to the olution variable of the reduced form; in thi cae the olution variable of (5.3.) i the variable x. To interpret the tate model, we need to relate it tate variable x and x 2 to x. The value of the matrice C and D tell u that the output variable i y =.4x 2. Becaue the output y i the ame a x, we then ee that x 2 = x/.4 = 2.5x. The other tatevariable x i related to x 2 by the econd tate equation ẋ 2 = x. Thu, x = 2.5ẋ THE tfdata FUNCTION To create a tranfer function decription of the ytem y3, previouly created from the tate model, you type tfy3 = tf(y3). However, there can be ituation where we are given the model tfy3 in tranfer function form and we need to obtain the numerator and denominator. To extract and ave the coefficient of the tranfer function, ue the tfdata function a follow. [num, den] = tfdata(tfy3, 'v'); The optional parameter 'v' tell MATLAB to return the coefficient a vector if there i only one tranfer function; otherwie, they are returned a cell array. For thi example, the vector returned are num = [,,.2] and den = [,.4,.8]. Thi correpond to the tranfer function X () F() = = which i the correct tranfer function, a een from (5.2.2). Tranfer Function of a TwoMa Sytem EXAMPLE 5.3. Problem Obtain the tranfer function X ()/F() and X 2 ()/F() of the tatevariable model obtained in Example The matrice and tate vector of the model are A = B = 3
25 274 CHAPTER 5 Block Diagram, StateVariable Model, and Simulation Method and x x z = x 2 x 3 = x 2 ẋ x 4 ẋ 2 Solution Becaue we want the tranfer function for x and x 2, we mut define the C and D matrice to indicate that z and z 3 are the output variable y and y 2. Thu, C = The MATLAB program i a follow. [ ] D = [ A = [,,, ;,,, ;... , 4/5, 2/5, 8/5; 4/3, 4/3, 8/3, 8/3]; B = [; ; ; /3]; C = [,,, ;,,, ]; D = [; ] y4 = (A, B, C, D); tfy4 = tf(y4) The reult diplayed on the creen are labeled # and #2. Thee correpond to the firt and econd tranfer function in order. The anwer are X () F() = X 2 () F() = ] Table 5.3. ummarize thee function. Table 5.3. LTI object function. Command Decription y = (A, B, C, D) Create an LTI object in tatepace form, where the matrice A, B, C, and D correpond to thoe in the model ẋ = Ax + Bu, y = Cx + Du. [A, B, C, D] = data(y) Extract the matrice A, B, C, and D of the LTI object y, correponding to thoe in the model ẋ = Ax + Bu, y = Cx + Du. y = tf(num,den) Create an LTI object in tranfer function form, where the vector num i the vector of coefficient of the tranfer function numerator, arranged in decending order, and den i the vector of coefficient of the denominator, alo arranged in decending order. y2=tf(y) Create the tranfer function model y2 from the tate model y. y=(y2) Create the tate model y from the tranfer function model y2. [num, den] = tfdata(y, 'v') Extract the coefficient of the numerator and denominator of the tranfer function model y. When the optional parameter 'v' i ued, if there i only one tranfer function, the coefficient are returned a vector rather than a cell array.
26 5.3 StateVariable Method with MATLAB LINEAR ODE SOLVERS The Control Sytem Toolbox provide everal olver for linear model. Thee olver are categorized by the type of input function they can accept: zero input, impule input, tep input, and a general input function THE initial FUNCTION The initial function compute and plot the free repone of a tate model. Thi i ometime called the initial condition repone or the undriven repone in the MATLAB documentation. The baic yntax i initial(y,x), where y i the LTI object in tate variable form, and x i the initial condition vector. The time pan and number of olution point are choen automatically. Free Repone of the TwoMa Model EXAMPLE Problem Compute the free repone x (t) and x 2 (t) of the tate model derived in Example 5.2.3, for x () = 5, ẋ () = 3, x 2 () =, and ẋ 2 () = 2. The model i ẋ = x 3 ẋ 2 = x 4 ẋ 3 = 5 ( 5x + 4x 2 2x 3 + 8x 4 ) ẋ 4 = 3 [4x 4x 2 + 8x 3 8x 4 + f (t)] or ẋ = Ax + B f (t) where and A = x x x = x 2 x 3 = x 2 ẋ x 4 ẋ 2 B = 3 Solution We mut firt relate the initial condition given in term of the original variable to the tate variable. From the definition of the tate vector x, we ee that x () = 5, x 2 () =, x 3 () = 3, x 4 () = 2. Next we mut define the model in tatevariable form. The ytem y4 created in Example 5.3. pecified two output, x and x 2. Becaue we want to obtain only one output here
27 276 CHAPTER 5 Block Diagram, StateVariable Model, and Simulation Method Figure 5.3. Repone for Example plotted with the initial function. Figure Repone for Example plotted with the plot function. Amplitude To: Out(2) To: Out() 5 4 Repone to Initial Condition Time (econd) 3 x 2 x t (x ), we mut create a new tate model uing the ame value for the A and B matrice, but now uing [ ] [ ] C = D = The MATLAB program i a follow. A = [,,, ;,,, ;... , 4/5, 2/5, 8/5; 4/3, 4/3, 8/3, 8/3]; B = [; ; ; /3]; C = [,,, ;,,, ]; D = [; ] y5 = (A, B, C, D); initial(y5, [5,, 3, 2]) The plot of x (t) and x 2 (t) will be diplayed on the creen (ee Figure 5.3.). To plot x and x 2 on the ame plot you can replace the lat line with the following two line. [y,t] = initial(y,[5,,3,2]); plot(t,y),gtext('x '),gtext('x 2'),xlabel('t') The reulting plot i hown in Figure To pecify the final time tfinal, ue the yntax initial(y,x,tfinal). To pecify a vector of time of the form t = (:dt:tfinal), at which to obtain the
28 5.3 StateVariable Method with MATLAB 277 olution, ue the yntax initial(y,x,t). When called with lefthand argument, a [y, t, x] = initial(y,x,...), the function return the output repone y, the time vector t ued for the imulation, and the tate vector x evaluated at thoe time. The column of the matrice y and x are the output and the tate, repectively. The number of row in y and x equal length(t). No plot i drawn. The yntax initial(y,y2,...,x,t) plot the free repone of multiple LTI ytem on a ingle plot. The time vector t i optional. You can pecify line color, line tyle, and marker for each ytem; for example, initial(y,'r',y2, 'y',y3,'gx',x) THE impule, tep, AND lim FUNCTIONS You may ue the impule, tep, and lim function with tatevariable model the ame way they are ued with tranfer function model. However, when ued with tatevariable model, there are ome additional feature available, which we illutrate with the tep function. When called with lefthand argument, a [y, t] = tep(y,...), the function return the output repone y and the time vector t ued for the imulation. No plot i drawn. The array y i (p q m), where p i length(t), q i the number of output, and m i the number of input. To obtain the tate vector olution for tatepace model, ue the yntax [y, t, x] = tep(y,...). To ue the lim function for nonzero initial condition with a tatepace model, ue the yntax lim(y,u,t,x). The initial condition vector x i needed only if the initial condition are nonzero. Thee function are ummarized in Table Table Baic yntax of linear olver for tate variable model. initial(y,x,tfinal) Generate a plot of the free repone of the tate variable model y, for the initial condition pecified in the array x. The final time tfinal i optional. initial(y,x,t) Generate the free repone plot uing the uerupplied array of regularlypaced time value t. [y,t,x]=initial(y,x,...) Generate and ave the free repone in the array y of the output variable, and in the array x of the tate variable. No plot i produced. tep(y) Generate a plot of the unit tep repone of the LTI model y. tep(y,t) Generate a plot of the unit tep repone uing the uerupplied array of regularlypaced time value t. [y,t]= tep(y) Generate and ave the unit tep repone in the array y and t. No plot i produced. [y,t,x]=tep(y,...) Generate and ave the free repone in the array y of the output variable, and in the array x of the tate variable, which i optional. No plot i produced. impule(y) Generate and plot the unit impule repone of the LTI model y. The extended yntax i identical to that of the tep function. lim(y,u,t,x) Generate a plot of the total repone of the tate variable model y. The array u contain the value of the forcing function, which mut have the ame number of value a the regularlypaced time value in the array t. The initial condition are pecified in the array x, which i optional if the initial condition are zero. [y,x]= lim(y,u,t,x) Generate and ave the total repone in the array y of the output variable, and in the array x of the tate variable, which i optional. No plot i produced.
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